LESSON 2: CIRCLES AND SECTORS

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Deborah Carr
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1 LESSON : CIRCLES AND SECTORS G.C.1 1. B similar G.C.1. Similar figures have the same shape and proportional size differences. This is true of circles in which the radius is used to scale the figure larger or smaller. G.C units G.C units G.C cm G.C cm G.C cm Challenge Problem G.C.5. a. The surface area of the cone without the base is equal to the area of the remaining of the circle. 3 π in b. The circumference of the base of the cone will equal of the circle circumference ( π). in. You can use the circumference to determine the cone base diameter and radius in. π So, the diameter is in. and the radius is in. c. Since ou have the cone base radius, ou can calculate the cone base area. π 50. in Copright 015 Pearson Education, Inc.

2 LESSON 3: DEGREES AND RADIANS G.C.5 1. C π radians G.C.5. B π radians G.C.5 3. D 10 G.C.5. π or 0.79 radians G.C π or.3 radians G.C G.C G.C radians Challenge Problem G.C.5 9. θ= = s r radians Copright 015 Pearson Education, Inc. 9

3 LESSON : AREA OF A SECTOR G.C.5 1. C 1.13 m G.C.5. B. m G.C cm G.C in G.C cm G.C.5..3 m G.C.5 7. The total pizza area is π in. Challenge Problem One piece is a sector with a central angle of 7. The surface area of each piece is in. G.C.5. The total pizza area is π( in.) = 00.9 in. The larger piece has a central angle of 5, so the surface area is in. 30 One of the smaller pieces has a surface area of in. 30 The larger piece has about 3.3 in more surface area than one of the smaller pieces. Copright 015 Pearson Education, Inc. 50

4 LESSON 5: ANGLES IN CIRCLES G.C. 1. D 30 G.C.. C 90 G.C. 3. A 7 G.C.. B 0 G.C. 5. A 5 G.C.. B 50 G.C. 7. D 10 G.C.. C 50 Challenge Problem G.C. 9. Here are the steps I followed to find the measure of CFE. I know that DEF is a right angle since DF is the diameter. Therefore, based on DEF, ou can find the measure of CFE = 3. A D C B F Here is another method to determine the measure of CFE. ABD forms a straight angle with DBE, so the two angles are supplementar. Thus, the measure of DBE = 10. You know the measures of two of the three angles for DBE, so ou can determine the measure of the third angle: the measure of BED = 9. E You know the measure of DEF = 90, since it spans a diameter. Thus, BED and BEF are complementar, so ou can compute the measure of BEF = 1. FBE is vertical with ABD, so the two angles are equal in measure: the measure of FBE = 7. Now ou know the measures of two of the three angles of BFE, so ou can compute the measure of BFE = 3. BFE and CFE are the same angle since points C and B are both on the same diameter. So, the measure of CFE = 3º. Copright 015 Pearson Education, Inc. 51

5 LESSON : CHORDS AND TANGENTS G.C. 1. CAD = 90º G.C.. ACD = 0º G.C. 3. B 35 G.C.. D 13 units G.C. 5. A 17 units G.C.. BCA = 70º G.C. 7. FG is 10 cm long. G.C.. ADC = 130º Challenge Problem G.C. 9. Here are the steps I followed to find the measure of FEC. As I found the measures of new angles, I recorded them in this diagram. G E F C DBC makes a straight angle with the given 10 angle, so the measure of DBC = 0. CDB is a right angle since DB is tangent at point D. I now have two of the three angle measures for CBD. Therefore, the measure of DCB = 50. DCB is vertical with ECF, so the angle measures are equal: ECF = 50. CE is congruent to CF since both are radii of the same circle, so CEF is isosceles. Thus, the measures of FEC and EFC must be equal. Since I know ECF = 50, then the measures of FEC and EFC must each be half of 130, or 5. Therefore, FEC = 5. D 10 0 B Copright 015 Pearson Education, Inc. 5

6 LESSON 7: CIRCLES AND QUADRILATERALS G.C.3 1. B cclic G.C.3. A supplementar (add up to 10 ) G.C.3 3. ADC = 7 G.C.3. BDA = 30 G.C.3 5. To find the measure of BDA, I started b finding the measure of the third angle in ABC. 10 = BCA BCA = 30º G.C.3. DAP = 50º G.C.3 7. D isosceles G.C.3. B 9 G.C.3 9. BPD = 150 Challenge Problem BCA BDA because of the theorem stating that if two inscribed angles of a circle intercept the same arc or arcs of equal measure, then the inscribed angles have equal measure. Therefore, BDA = 30º. G.C A 0 The diagonals intersect and make vertical angles, so ou can fill in all of the angle measures around that intersection. With this information, ou can then fill in all of the other angle measures shown, using the fact that the sum of the angle measures of a triangle add up to E In order to show that the quadrilateral 9 70 B C 110 is cclic, ou need to show that opposite 39 angles add up to 10. CAE = and CDE = 11 : + 11 = ACD = 1 and AED = 99 : = 10 D Both pairs of opposite angle measures of the quadrilateral add up to 10. Therefore, the quadrilateral is cclic. Copright 015 Pearson Education, Inc. 53

7 LESSON : DESCRIBING CIRCLES G.GPE.1 1. B ( 3, ) G.GPE.1. B 5 units G.GMD. 3. C Parallel to the base of the cone G.GPE.1. A (x + 3) + ( ) = 9 G.GPE.1 5. The center is at (, ). G.GPE.1. The radius = units G.GPE (, ) 10 x G.GPE.1. The center is at (3, 7). G.GPE.1 9. The radius = 7 units G.GPE (3, 7) 7 10 x Copright 015 Pearson Education, Inc. 5

8 LESSON : DESCRIBING CIRCLES G.GPE The center is at (, ). G.GPE.1 1. The radius = 5 units G.GPE (, ) 5 10 x Challenge Problem G.GPE.1 1. a. This equation is different from the equation of a circle because the two denominators are different. Instead of indicating the same radius in both horizontal and vertical directions, this equation indicates different radii horizontall and verticall. b. This equation describes an ellipse with a center at (3, ). c. 10 x Copright 015 Pearson Education, Inc. 55

9 LESSON 9: CIRCLE: AS AN EQUATION G.GPE.1 1. Center: (3, 5) Radius: units G.GPE.1. B x + 1x = 0 E x G.GPE.1 3. (x + ) + ( + ) = 5 G.GPE.1. D x 10x + + = 0 G.GPE.1 5. (x + 1) + ( ) = 1 x + x = 1 x + x = 0 G.GPE.1. x + + x = 0 x + x + 1 = 1 x + x = (x + 3) + ( ) = G.GPE.1 7. x + 1x + + = 0 x 1x + + = x 1x = (x 7) + ( + ) = 5 G.GPE.1. Center: (, ) Radius: units Copright 015 Pearson Education, Inc. 5

10 LESSON 9: CIRCLE: AS AN EQUATION G.GPE x (, ) 10 G.GPE Center: (, 9) Radius: units G.GPE (, 9) x Challenge Problem G.GPE.1 1. This equation does not represent a circle. You can tell just from the equation, because it has the 3x term. Because of this term, the equation will not simplif into an equation of a circle after completing the square. Here is the graph of this equation. It makes a rotated ellipse. x + 3x + + 1x + 17 = x 0 30 Copright 015 Pearson Education, Inc. 57

11 LESSON 10: DESCRIBING PARABOLAS G.GMD. 1. A Parabola G.GPE.. B directrix, focus G.GPE. 3. C point C G.GPE.. The directrix is = 5, or line l. G.GPE. 5. D FO and DO are the same length. G.GPE.. The coordinates of the focus are (0, 0). G.GPE. 7. The focus is at (0, ) and the directrix is = = x Focus: (0, ) x Directrix: = Copright 015 Pearson Education, Inc. 5

12 LESSON 10: DESCRIBING PARABOLAS G.GPE.. The focus is at (0, 5.5) and the directrix is = x 3 1 = x 5 Focus: (0, 5.5) 7 Directrix: =.5 Challenge Problem G.GPE. 9. The focus is at (0, 17 ) and the directrix is = Copright 015 Pearson Education, Inc. 59

13 LESSON 11: PARABOLA: AS AN EQUATION G.GPE. 1. A 1 = x + G.GPE.. = 1 x G.GPE. 3. G.GPE.. 1 x = = x G.GPE. 5. B Focus at (0, ) and directrix of = G.GPE.. The focus is at (3, 0). The directrix is x = 1. G.GPE The focus is at 0,. The directrix is = Challenge Problem G.GPE.. This equation describes a sidewas parabola with vertex at (, ) x Copright 015 Pearson Education, Inc. 0

14 LESSON 1: CIRCLES AND PARABOLAS G.GMD. G.MG.1 G.GMD. G.MG.1 G.GMD. G.MG.1 G.GMD. G.MG.1 1. C = π cm A = 1π cm. C = 1π cm A = π cm 3. C = π cm A = 1π cm. The circumference decreases proportionall with how far up the circle the cross section is. In this case, the circumference of the base is 75. cm, the circumference of the middle is 50. cm, and the circumference of the top is 5.13 cm (or π cm, 1 1 π cm, and π cm), which follows the 1,, pattern of distances from the top of the cone. 3 3 The area changes from 1π cm to π cm to 1π cm. These areas do not follow a proportional pattern: 1 π π = 5. but = π 1π The pattern is derived from the ratios squared. The middle area is = of the base area and the top area is = of the base area. 3 9 G.GPE.1 5. This sketch is of a cross section of the scenario. Since the parabola must be parallel with the side of the cone, ou can determine that these triangles are similar A 0 B x Copright 015 Pearson Education, Inc. 1

15 LESSON 1: CIRCLES AND PARABOLAS G.GPE.1. Because the triangles are similar, the length of AB is proportional to the large triangle created from the cone. The slant height of the cone can be computed using the Pthagorean Theorem = = The length of AB is cm. G.GPE.1 7. x + ( 7.5) = 5.5 Challenge Problem G.GMD.. The different conic sections are determined b the angle at which the plane intersects the double cone. The circle onl occurs when the plane is parallel to the base of the double cone. The parabola onl occurs if the plane is parallel to the side of the cone. And the hperbola occurs when the plane is at a steeper angle than parallel to the side of the cone (closer to perpendicular to the base of the cone). Copright 015 Pearson Education, Inc.

16 LESSON 13: PUTTING IT TOGETHER G.C.. Word of Phrase Definition Example Central angle An angle with its vertex at the center of the circle and two radii as its sides You can measure central angles with degrees or radians. Arc length s = radius r s r = 1 radian Circle Conic Section Definition: The intersection of a right circular cone with a plane such that the plane is perpendicular to the altitude of the cone Points Definition: A collection of points on a plane that have a fixed distance r (the radius) from a fixed point C (the center) in the plane Sectors A region bounded b two radii and an arc of a circle s r Arc Length The distance along an arc for a particular angle or portion of the circumference of a circle The arc length of the above sector is s = r θ (in radians). (continues) Copright 015 Pearson Education, Inc. 3

17 LESSON 13: PUTTING IT TOGETHER G.C.. (continued) Word of Phrase Definition Example Chord A line segment with a circle that has its end points on the circle MN is a chord. T S P M O N Q Diameter The longest chord of a circle B x A Tangent A line that intersects a circle at just one point PS is a tangent. S T P M O N Q (continues) Copright 015 Pearson Education, Inc.

18 LESSON 13: PUTTING IT TOGETHER G.C.. (continued) Word of Phrase Definition Example Secant A line that intersects a circle at two points PQ is a secant. S T P M O N Q Copright 015 Pearson Education, Inc. 5

19 LESSON 13: PUTTING IT TOGETHER G.C. 3. Theorem (Lesson Number) Definition Explain the Theorem Using Words or Diagrams Arc Length-Radius Theorem (Lesson 3) In sectors with a common central angle θ, the arc length of the sector is proportional to the radius. The arc length of this sector is determined b both the radius and the central angle. s = r θ (in radians) s r Area of a Sector of a Circle (Lesson ) The area of a sector with central angle θ in degrees is θ A =πr. 30 The area of a sector with central angle θ in radians is r θ A =. The radius is squared and then multiplied b pi and the portion of the circle that the central angle sweeps. If the sector is 10º, or 7π, and the radius is 3, then the area of the sector is 9 7 π = 55. π units. When calculated with the degree formula, π 9 10 = 55. π units. 30 Inscribed Angle Theorem (Lesson 5) An inscribed angle of a circle has a measure exactl half of the angle measure of the intercepted arc. n C b m A n x a m B In this diagram, a = b because b is an inscribed angle that intercepts arc AB while a is the central angle. (continues) Copright 015 Pearson Education, Inc.

20 LESSON 13: PUTTING IT TOGETHER Copright 015 Pearson Education, Inc. 7

21 LESSON 13: PUTTING IT TOGETHER EXERCISES G.C. 3. (continued) Theorem (Lesson Number) Definition Explain the Theorem Using Words or Diagrams Tangents From an External Point Theorem (Lesson ) Tangent segments to a circle from the same external point are congruent. P T O S Circumscribed Angle Theorem (Lesson ) When a circumscribed angle of a circle shares a common chord with the central angle, the two angles are supplementar. PT is congruent to PS. In the above tangents diagram, TPS is supplementar to T0S because the share the common chord TS. Cclic Quadrilateral Angles Theorem (Lesson 7) A quadrilateral is cclic if and onl if its opposite angles are supplementar. D F C B A ABC + ADC = 10º and BAD + BCD = 10º Copright 015 Pearson Education, Inc.

22 LESSON 13: PUTTING IT TOGETHER G.GPE.1 G.GPE. G.GMD.. Circle Parabola 5 Focus: (0, ) 3 (, 3) Directrix: = x x = 1 1 x General Formula for a Circle With a center at (h, k) and a radius of r, the circle has the formula (x h) + ( k) = r. In polnomial form, it is x + Dx + + E + F = 0, which can be converted to center-radius form b completing the square. General Formula for a Parabola With a focus at (0, p) and a directrix of = p and the parabola opening up or down, the parabola has the formula = 1 p x. If the parabola opens left or right, then x and are switched such that the -term is squared and the x-term is not squared. Circle as a Conic Section The intersection of a right circular cone with a plane such that the plane is perpendicular to the altitude of the cone. Parabola as a Conic Section The intersection of a right circular cone with a plane such that the plane is parallel to the side of the cone. Copright 015 Pearson Education, Inc. 9

Understand and Apply Theorems about Circles

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